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I'm just starting out in Computer Science and currently learning about transmission of data in fibre-optic cables. I understand fibre-optic attenuation and that a repeater will then take this light and amplify it.

My question is, how is the repeater doing this? What is the process in which the repeater takes light in and converts this to binary code? I completely understand the physical pulses represent 1 and 0, but how is this being read by the repeater?

To my very little understanding, I'm assuming the light is not pulsing any electrical current of any kind to be read by transistors like how they would be in a physical computer.

Thanks.

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    \$\begingroup\$ A repeater is usually just a combination of receiver and transmitter. It has its own power source of course. \$\endgroup\$
    – Eugene Sh.
    Commented Jan 8, 2021 at 14:24
  • \$\begingroup\$ Light is still energy; we can convert between light and electrical (and vice versa). For most short range (within a data centre for instance) fibre channel setups the transmitter is a VCSEL and the receiver is a PIN diode, the output of which is amplified. \$\endgroup\$ Commented Jan 8, 2021 at 14:28
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    \$\begingroup\$ What does "binary" mean to you? \$\endgroup\$ Commented Jan 8, 2021 at 17:19
  • \$\begingroup\$ Required reading about modern 100 Gigabit optical transceivers and how incredibly clever they are: ciena.com/insights/articles/… \$\endgroup\$
    – tomnexus
    Commented Jan 8, 2021 at 19:36

3 Answers 3

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How does a repeater convert light from a fibre-optic cable into binary?

The received light is converted to an electrical signal using (typically) a photodiode: -

enter image description here

It converts incident light power into a current. The current is (usually) converted to a voltage using a TIA (transimpedance amplifier): -

enter image description here

And the voltage can be converted into a series of logic bits using a voltage comparator: -

enter image description here

That's the simple story.

The more complex story involves automatic gain circuits in case the light level changes over time and distance (usually the case). The more complex story can involve a tweaking of the hardware based on what data is received i.e. if error rate appears to be a little too high, there can be feedback to the hardware that tweaks things this way or that way in order to reduce the error rate.

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  • \$\begingroup\$ Ah, I've never heard of a photodiode before or the process of light being converted into a current. That completely explains it all. Thank you very much, Andy. \$\endgroup\$
    – BuiltByDan
    Commented Jan 8, 2021 at 14:36
  • \$\begingroup\$ Is it photodiodes rather than phototransistors because of the high speed used, or for some other reason (cost etc)? \$\endgroup\$
    – Lundin
    Commented Jan 8, 2021 at 14:40
  • \$\begingroup\$ @BuiltByDan Photodiodes/phototransistors are everywhere: TV remotes, garage obstacle sensors, automatic doors, light sensitive night lights, mechanical mice (not that these are common these days though) and what not.. \$\endgroup\$
    – Eugene Sh.
    Commented Jan 8, 2021 at 14:47
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    \$\begingroup\$ @Lundin it's down to speed and noise that we choose photodiodes. \$\endgroup\$
    – Andy aka
    Commented Jan 8, 2021 at 14:47
  • \$\begingroup\$ This is not correct. This describes a regenerator, that does an optical-electrical-optical conversion - re-generating the signal. A repeater stays in the optical domain, and only amplifies the signal optically, often using a Erbium-doped fiber amplifiers (EDFA). \$\endgroup\$
    – RJR
    Commented Jan 11, 2021 at 9:09
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To add to Andy's good answer:

That's the easiest case, and what you'll see on "low-speed fiberoptics", e.g. 1 gigabit and 10 gigabit per second transceivers.

It's called "intensity modulation/direct detection", and is really that: a laser diode modulates (i.e. changes) the intensity of light, and the receiver directly detects that with a photo diode.

You can do that with only 2 states (on / off), and still get pretty decent speed because light over fiber has such a high bandwidth in which things basically behave the same.

Soon as that breaks, you will need to either add more levels (e.g. off / ¼ on / half on / full on), which makes detection way harder, and increases the number of symbols that are "misdetected" and hence increases the need for error correction, or you need to be even cleverer.

You can do coherent receivers (and transmitters) on light, too. What that means is that you'd very much be modulating both the amplitude of light, as well as the phase of the wave. Demodulation then becomes quite a bit more tricky, and basically all the knowledge that you get about wireless communication applies, too. This takes a full specialization in electrical engineering, so I doubt you want to dig that deep this early in your studies – it is, in fact, a very interesting field for the math kind of computer scientists, too, as "we" do care about a lot of information theory, and questions like "what is the theoretical maximum capacity that we can transport over these non-linear channel", which involves a lot of math that we don't have yet; unlike other channels, where we know how much bits per second you can transport if you accept a certain probability of error (say, 10⁻¹⁴), we don't actually know that for the models of the fiberoptical channel that we use to design transceiver systems. Good fun!


What all this strives to show: the "computer science" approximation to reality that bits are bits is not true for the physical world. We always need to convert our bits to some physical entity – voltage, current, light intensity, phase of a radio wave, charge in a flash memory cell, air pressure … – and convert it back when reading it. This is always a lossy process, and there's a lot of technology behind making sure that the abstraction "I put a bit into and get the same bit out" works – and that it works under limitations!

For example, in most cases, you can make your system more reliable by putting more energy into every bit – but in case of phones, for example, that energy directly comes from the battery, and you don't want that to drain in half an hour just because a user is fetching cat pictures from a webserver.

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Using the correct terminlogy in Optical communications, a repeater is an optical amplifier that amplifies the optical signal, most often using an Erbium-doped optical fiber amplifiers (EDFA). It does not improve the signals quality per-se, it only increases it's magnitude allowing it to be transmitted over longer distances.

Conversely, a regenerator converts the optical signal back to an electrical signal, reconsitutes the bit steam(s) and converts them back to the optical domain, thereby creating a new signal relatively free of defects.

The advantage of repeating over regenerating is that for optical signals that contain multiple wavelengths (as used in Wave Division Multiplexing, or WDM), the signal can be amplified as-is, while a regenerator will have to first split the signal into is constituents wavelengths and regenerate each signal seperately, re-combining them into a single optical signal before it's transmitted.

In summary, you're incorrect in assuming a repeater does any optical to electrical conversion at all.

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    \$\begingroup\$ The question may have said "repeater" in its title but it also asked how it produces a binary code. Given that this is an EE site, there is a big hint here; an electrical interface is needed (and implied by the question) to convert the light. I have noted your downvote. \$\endgroup\$
    – Andy aka
    Commented Jan 11, 2021 at 9:28
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    \$\begingroup\$ OP: "I understand fibre-optic attenuation and that a repeater will then take this light and amplify it. My question is, how is the repeater doing this?" --> the title of the question indicates that OP does not understand what a repeater does. Explaining how OEO converson works is nice, but it doesn't help OP understand that his assumption of how a repeater works is incorrect. It does NOT convert optical to electrical and back again. \$\endgroup\$
    – RJR
    Commented Jan 11, 2021 at 12:38
  • \$\begingroup\$ You appear to be saying that an optical repeater never uses a process of converting to electrical then re-transmitting. And that is wrong. \$\endgroup\$
    – Andy aka
    Commented Jan 11, 2021 at 12:43
  • \$\begingroup\$ @Andyaka you are wrong here - there are optical amplifiers available for amplfication "underway" in long-haul optical communications, which amplify the light as light – no conversion to an electrical signal necessary. EDFA is really the keyword here! \$\endgroup\$ Commented Jan 11, 2021 at 13:04
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    \$\begingroup\$ @MarcusMüller I'm not saying that optical amplifiers don't exist; read again what I wrote and note the reasons why I'm saying it (the OP is asserting that a repeater isn't what I described under my answer). Please also note the links I gave that DO define an optical repeater as covering (at least) a process I described. \$\endgroup\$
    – Andy aka
    Commented Jan 11, 2021 at 13:11

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